Not sure if this is research level? While analyzing a particular algorithm, I came across the following series:
$\sum_{k=1}^{m-1}\frac{3^{k}}{\prod_{i=1}^{k-1}2^{2^{i}}}$

Is there a closed form expression that illuminates asymptotic values in terms of $m$ accurately? In particular, I am interested in how slow this function grows compared to $\frac{3^{m}-1}{2}$ (which is the summation if one sets the denominator in each of the sum term to $1$).

Possibly is there a function that approximates $f(m)$ where $f(m)$ is such that $S_{m} =(\frac{3^{m}-1}{2})^{f(m)}$?

For large enough k, the denominator can be rewritten more simply. The result converges quickly, but otherwise gives me no clue as to a closed form. Gerhard "Ask Me About System Design" Paseman, 2012.06.14
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Gerhard PasemanJun 14 '12 at 22:34

1 Answer
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Presumably the sum $S_m$ can be given in a more compact way as
$S_m = 4\sum_{k=1}^{m-1} \frac{3^k} {2^{2^k}}.$ The limit should be a transcendental number, since the sum is extremely lacunary, so it seems unlikely that a closed for exists (since most closed form transcendental numbers are "named" constants").

$\sum_{n=0}^{\infty}(1/n!)$ is a lacunary sum with a transcendental value, yet it has what would generally be accepted as a closed form; I'm not objecting to your conclusion, just to the word, "so".
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Gerry MyersonJun 14 '12 at 22:56

Hi Igor, I am only interested in something that grows as fast as $S_{m}$ and not S_{m}$ accurately.
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TurboJun 14 '12 at 23:51

The sequence $S_m$ converges to some number between $2.83$ and $2.84$, so it doesn't grow very much...
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François G. Dorais♦Jun 15 '12 at 0:23